Another way to look at it is to think that a square root is also x^½ (x raised to the half power)

For example √x² = x is is the same as saying (x²)^½ = x^(2/2) = x^1 = x

So, 4√x³y x 4√xy² = x^(3/4) × y^(1/4) × x^(1/4) × y^(2/4)

Combining terms: x^(4/4) × y^(3/4)

Simplifying: x × y^(3/4) = x 4√y³ (x times fourth root of y cubed)

mathsyperson

2005-07-23 04:06:04

√a x √b= √(a x b), so 4√x³y x 4√xy² = 4√(x³y x xy²)

Multiply the terms: 4√((x^4)y³) 4√x^4=x, so taking x out of the 4th root gives x(4√y³)

Your answer should have the 4th root of y³ instead of the square root. So, yes, minor typo error.

samsoo

2005-07-23 03:48:50

One more thing, is this a typo error agian or correct answer?

samsoo

2005-07-23 03:43:11

Hi thx for the reply.

You're right, it does work if there's a plus sign between 3/2(x-1) and √(2x²-7x-4). I checked the question and there's no plus sign, so I guess it could be a typo error.

I was given a typed up booklet just to familiarise and practise on, I hope there's no more other typos or I'll just get stressed up for doing wrong questions.

mathsyperson

2005-07-23 01:04:03

I tried working backwards by squaring the answer, and found that you actually want to find to square root of:

3/2(x-1)+√(2x²-7x-4)

I wouldn't have any idea how to solve it if you hadn't already given us the answer though. Squaring is easy, you can just follow a set of rules to get the answer every time. Isn't it weird how the inverse can be so much more horrible?

samsoo

2005-07-22 23:44:01

Hi all.

I'm new here and need some help with this maths problem.

Here's a pic of the question and the answer and I've tried doing a method but my head can't get round this.